IIT JEE Advanced 2007 Paper 2

We have

dy

dx

=

√1−y2

y

⇒

y

√1−y2

dy = dx
On integrating both sides, we get
∫

ydy

√1−y2

= ∫ dx
⇒ ∫

−2ydy

√1−y2

= - 2 ∫ dx
⇒ 2√1−y2 = - 2x + 2c
⇒ √1−y2 = - x + c
⇒ 1−y2 = (x−c)2
⇒ (x−c)2+y2 = 1
which is the equation of circle with fixed radius 1 and variable centre along x-axis.